CSCI 2670 Introduction to Theory of Computing November 1, 2005

Agenda Last week Today Test & undecidability Review one undecidability proof Rice’s Theorem Reductions (Section 5.3) November 1, 2005

Announcements Announcements Homework due next Tuesday (11/8) 5.7, 5.12 (do not use Rice’s Theorem), 5.20, 5.22, 5.30 b Old text 5.7 Get handout for 5.12, 5.20, 5.22 and 5.30 b November 1, 2005

An undecidable language Let REGULARTM = {<M> | M is a TM and L(M) is a regular language} Theorem: REGULARTM is undecidable Proof: Assume R decides REGULARTM and use R to decide ATM (reduce the ATM problem to the REGULARTM problem). As before, make a new TM, M2, that accepts a regular language iff M accepts w. November 1, 2005

Proof (cont.) Consider the following TM S = “On input <M,w> Construct the following TM M2 M2 = “On input x If x = 0n1n for some n, accept Otherwise, run M on w. If M accepts w, accept” Run R on M2 (accepts iff L(M2) is a RL) If R accepts, accept; if R rejects, reject” S decides ATM if R decides REGULARTM L(M2) = 0n1n if M does not accept w L(M2) = all strings if M accepts w November 1, 2005

Insight The TM M2 is specially designed to be regular if and only if M accepts w Then call TM that decides REGULARTM on M2 November 1, 2005

Rice’s theorem Determining whether a TM satisfies any non-trivial property is undecidable A property is non-trivial if: It depends only on the language of M, and Some, but not all, Turing machines have the property Examples: Is L(M) regular? A CFG? Finite? November 1, 2005

Proof of Rice’s theorem Assume there is some decidable non-trivial property P for Turing machines Assume TM’s that accept  do not satisfy P If they do, just consider P November 1, 2005

Proof of Rice’s theorem Let TM B decide P On input <M>, B accepts iff TM M has property P Let MP be a TM that satisfies P Since P is non-trivial, there is some MP satisfying P Use B and MP to decide ATM Create a new TM S that decides ATM using B and MP November 1, 2005

ATM decider using MP S = “On input <M,w> Create the following TM N N = “On input x Run M on w until it accepts If M accepts w, run MP on x If MP accepts x, accept; if MP rejects x, reject” Run B on <N(<M,w>)> If B accepts, accept; if B rejects, reject” L(N) = L(MP) if M accepts w; otherwise L(N) =  S decides ATM if B decides TM’s satisfying P Therefore, B cannot exist November 1, 2005